AMC10 2012 A

AMC10 2012 A · Q11

AMC10 2012 A · Q11. It mainly tests Triangles (properties), Ratios in geometry.

Externally tangent circles with centers at points A and B have radii of lengths 5 and 3, respectively. A line externally tangent to both circles intersects ray AB at point C. What is BC?

中心在点 A 和 B 的外部相切圆分别有半径 5 和 3。一条与两个圆都外部相切的直线与射线 AB 相交于点 C。求 BC 的长度。

(A) 4 4

(B) 4.8 4.8

(D) 12 12

(E) 14.4 14.4

Answer

Correct choice: (D)

正确答案：（D）

Solution

Answer (D): Let $D$ and $E$ be the points of tangency to circles $A$ and $B$, respectively, of the common tangent line that intersects ray $AB$ at point $C$. Then $AD=5$, $BE=3$, and $AB=5+3=8$. Because right triangles $ADC$ and $BEC$ are similar, it follows that \[ \frac{BC}{AC}=\frac{BE}{AD}, \] so \[ \frac{BC}{BC+8}=\frac{3}{5}. \] Solving gives $BC=12$.

答案（D）：设$D$和$E$分别是公切线与以$A$、$B$为圆心的圆的切点，该公切线与射线$AB$相交于点$C$。则$AD=5$，$BE=3$，且$AB=5+3=8$。因为直角三角形$ADC$与$BEC$相似，所以 \[ \frac{BC}{AC}=\frac{BE}{AD}, \] 因此 \[ \frac{BC}{BC+8}=\frac{3}{5}. \] 解得$BC=12$。

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